Question 119577
Solve by the substitution method
3d+e over 4 = d+1 over 2
d-e over 4 = 1
;
{{{((3d+e))/4}}} = {{{((d+1))/2}}}
Cross multiply and you have:
2(3d+e) 4(d+1)
6d + 2e = 4d + 4
6d - 4d + 2e = 4
2d + 2e = 4
Simplify, divide by 2:
d + e = 2
e = (2-d); Use this for substitution 
:
{{{((d-e))/4}}} = 1 
Multiply equation by 4 to get rid of that annoying denominator, we have:
d - e = 4
:
Substitute (2-d) for e in the above equation:
d - (2 - d) = 4
d - 2 + d = 4
2d = 4 + 2
2d = 6
d = 3
:
Find e substituting 3 for d in equations; e = (2 - d)
e = 2 - 3
e = -1
:
:
Check solutions in the original 1st equation:
{{{((3d+e))/4}}} = {{{((d+1))/2}}}
{{{((3(3)-1))/4}}} = {{{((3+1))/2}}}
{{{((8))/4}}} = {{{((4))/2}}}; confirms our solutions
:
You can check solution in the original 2nd equation